What can be said about the ring $\mathbb{C}[x,y]/\langle xy \rangle$?

I was very certain that $$\mathbb{C}[x,y]/\langle xy \rangle \cong\mathbb C[x] \oplus\mathbb C[y]$$ since the elements in $\mathbb{C}[x,y]/\langle xy \rangle$ are of the form $$a_0+\sum_{i=1}^{n}{b_ix^i}+\sum_{i=1}^{m}{c_iy^i}$$ and from that I deduced the above isomorphism.

While I am still pretty sure that it is true, I cannot formulate a proof. So I am asking, is this true? If not, can someone show why? and maybe give another interpretation for this ring?

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    $\begingroup$ Well, it's not a direct sum, since constants lie in $\Bbb C[x]$ and also in $\Bbb C[y]$... Something very close to what you said is true (and being true may make it easier to prove). $\endgroup$ – David C. Ullrich Aug 29 '15 at 14:36
  • $\begingroup$ I originally thought that you can somehow "split" the constants and make it work. How can I ignore constants? $\endgroup$ – Mike Aug 29 '15 at 14:46
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    $\begingroup$ Spltting the constants shows that in fact $\Bbb C[x,y] = \Bbb C[x]+\Bbb C[y]$. But that's not a direct sum. For example there are many different ways to write $1=p+q$ with $p\in \Bbb C[x]$ and $q\in \Bbb C[y]$ (two that spring to mind are $1=1+0$ and $1=0+1$). You can't ignore the constants. But you could show, for example, that you have a direct sum $R\oplus S$ where, say, $R$ is the class of polynomials in $\Bbb C[x]$ with vanishing constant term and $S=\Bbb C[y]$. $\endgroup$ – David C. Ullrich Aug 29 '15 at 14:51
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    $\begingroup$ Ah. I was overlooking the fact that our isomorphism should preserve multiplication, sorry. Don't ask me to prove it, but I think Hoot must be right, the decomposition you want simply doesn't exist. $\endgroup$ – David C. Ullrich Aug 29 '15 at 15:11
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    $\begingroup$ Mike and @Hoot, it's not a nontrivial direct sum of rings with or without unity. "Without unity" doesn't give you any more freedom since if $R_1 \oplus R_2$ has unity, $R_1$ and $R_2$ both have unity. Anyway I've explained this all in my answer. $\endgroup$ – 6005 Aug 29 '15 at 17:49

$\mathbb{C}[x,y] / \langle xy \rangle$ is not isomorphic to $\mathbb{C}[x] \oplus \mathbb{C}[y]$ because the first ring has no idempotents other than $0,1$, while the second does.

In fact, $\mathbb{C}[x,y] / \langle xy \rangle$ cannot be written as a direct sum of any two nonzero rings:

In a commutative ring with unity $R$, if $R = R_1 \oplus R_2$, then there is an idempotent $e \in R$ such that $R_1 \cong eR$ and $R_2 \cong (1-e)R$. And conversely, if $e$ is an idempotent of $R$, then $R \cong eR \oplus (1-e)R$. (Here, $eR$ is the ideal generated by $R$, which forms a ring with identity $e$.) The special cases $R = 0 \oplus R$ and $R = R \oplus 0$ correspond to the idempotents $0, 1$ in $R$.

In $\mathbb{C}[x,y] / \langle xy \rangle$, $0$ and $1$ are the only idempotent elements. (If $P \in \mathbb{C}[x,y]$ has any powers of $x$ or powers of $y$, then $P^2$ has larger powers of $x$ or powers of $y$ and is not equal to $P$.) So $\mathbb{C}[x,y] / \langle xy \rangle$ can't be written as a direct sum $R_1 \oplus R_2$, except in the trivial way.

About rings without unity:

Can $R = \mathbb{C}[x,y]$ be written as $R_1 \oplus R_2$ if $R_1$ and $R_2$ don't need to be rings with unity? No, because then $R_1 \oplus R_2$ would have a unity element, say $(a,b)$, from which it follows that $a$ is a unity for $R_1$ and $b$ is a unity for $R_2$. So this does not allow any more flexibility.

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  • $\begingroup$ Thank you for you answer. It seems that other tools are needed when dealing with such rings. $\endgroup$ – Mike Aug 29 '15 at 17:56

The ring $\mathbb{C}[x,y]/(xy)$ cannot be a product of two rings since $\text{Spec }\mathbb{C}[x,y]/(xy) = \text{Spec }\mathbb{C}[x,y]/(x) \cup \text{Spec }\mathbb{C}[x,y]/(y)$ (from $(xy) = (x) \cap (y)$) and this is a union of two connected spaces whose intersection is nontrivial (the prime ideal $(x,y)$), hence connected.

It's a general fact that a commutative ring $A$ is a product iff $\text{Spec }A$ is disconnected (corrected). If $A$ is a product of two rings then it's clear that $\text{Spec }A$ is a disjoint union of open subsets, say $\text{Spec }A = V((0,1)) \cup V((1,0))$ if $A = A_1 \times A_2$ (because every prime ideal of the product $A_1 \times A_2$ is of the form either $\mathfrak{p} \times A_2$ or $A_1 \times \mathfrak{q}$, the former containing $(0,1)$ and the latter containing $(1,0)$) and for the converse, if $\text{Spec }A = U_1 \cup U_2$ with $U_1, U_2$ disjoint open subsets of $\text{Spec }A$, $A = \Gamma(\text{Spec }A, \mathcal{O}_X) = \Gamma(U_1, \mathcal{O}_X) \times \Gamma(U_2, \mathcal{O}_X)$ by the sheaf property, where $X = \text{Spec }A$.

Since $(x) \cap (y) = (xy)$, the natural map $\mathbb{C}[x,y]/(xy) \to \mathbb{C}[x] \oplus \mathbb{C}[y]$ is an injection, with the image $\{ (f(x),g(y)) \, | \, f(0) = g(0) \}$, i.e. you're glueing two lines at their origins.

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    $\begingroup$ Great geometric answer: +1. $\endgroup$ – Georges Elencwajg Sep 1 '15 at 7:55
  • $\begingroup$ @GeorgesElencwajg Thanks for showing me my cognitive mistake in my idea above: of course there at infinitely many deals in the quotient too. I suckered myself into thinking too much like a polynomial ring in a single variable. Now I see something like this is needed. Regards and thanks again $\endgroup$ – rschwieb Sep 1 '15 at 10:06
  • $\begingroup$ Dear @rschwieb : your gracious comment does you credit. Everybody makes little mistakes but not everybody reacts the gentlemanly way you did. $\endgroup$ – Georges Elencwajg Sep 1 '15 at 13:50

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